The next stage up from the quadratic equation, the general cubic equation has the form $ax^3+bx^2+cx+d=0.$

Just as there is a solution to the quadratic equation, a general solution to the cubic was found by Tartaglia and Ferro.

The problem is, even when there are three real solutions, the intermediate calculations use the roots of negative numbers, and this led directly to the acceptance of the complex numbers.

In practice, numerical solutions are found rather than using a closed form formula. Newton's Method is an efficient way to find solutions, although one must be aware that the basins of attraction have fractal boundaries.

Also related: quartic equation

## Outline of the solution

Start with a general cubic. Since the leading coefficient can't be zero we can divide through by that, so our general cubic is:
• $x^3+ax^2+bx+c=0$
Substitute $x=t-a/3$ giving
• $(t-a/3)^3+a(t-a/3)^2+b(t-a/3)+c=0$
Expand that and gather the t terms together and you get
• $t^3+pt+q=0$
for suitable p and q. Notice that there is no quadratic term.

Now substitute $t=u-\frac{p}{3u}$ and multiply by $u^3$ to get

• $u^6+qu^3+\frac{p^3}{27}=0$
Now we have a quadratic in $u^3$ so using the quadratic equation formula we get two possible answers for $u^3.$ Each of these gives three cube roots, giving six possible answers. These occur in pairs of identical solutions, giving three in all as required.